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Sample Problem: KCL 3

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Linear Circuits 1: DC Analysis

乔治亚理工学院

4.5（198 個評分） | 16K 名學生已註冊

This course explains how to analyze circuits that have direct current (DC) current or voltage sources. A DC source is one that is constant. Circuits with resistors, capacitors, and inductors are covered, both analytically and experimentally. Some practical applications in sensors are demonstrated.

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4.5（198 個評分）

5 stars
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4 stars
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3 stars
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2 stars
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1 star
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DM

May 09, 2018

Great course, learned a lot of things that they don't teach at my school, Thanks for the opportunity.

YZ

Jan 22, 2018

Excellent course where I get to learn a lot of things that is left from my AP Physics C E&M Course.

從本節課中

Module 2

The module describes some basic principles used in circuit analysis.

Sample Problem: DC Op Amp 19:29

Sample Problem: DC Op Amp 25:33

Sample Problem: DC Op Amp 34:23

Sample Problem: DC Op Amp 46:22

Sample Problem: DC Op Amp 55:33

Sample Problem: Current Division 13:57

Sample Problem: Current Division 25:20

Sample Problem: Parallel and Series Resistors 12:24

Sample Problem: Parallel and Series Resistors 28:07

Sample Problem: KVL6:05

Sample Problem: KVL 25:48

Sample Problem: KVL 32:28

Sample Problem: KCL 12:12

Sample Problem: KCL 24:43

Sample Problem: KCL 32:51

Sample Problem: KCL 41:26

Sample Problem: KCL 52:10

Sample Problem: Op Amps 19:17

Sample Problem: Op Amps 25:29

Sample Problem: Op Amps 34:59

Sample Problem: Nodes, Branches, Paths, and Loops3:19

Sample Problem: Summing Op Amp6:05

Sample Problem: Differential Amp6:47

Sample Problem: Inverting Op Amp5:09

Sample Problem: Non-inverting Op Amp7:54

Sample Problem: Max Power (Depend Sources)4:06

Sample Problem: Dependent Sources 15:10

Sample Problem: Dependent Sources 24:48

Sample Problem: Series/Parallel (Independ Sources) 12:59

Sample Problem: Series/Parallel (Independ Sources) 27:22

Sample Problem: Series/Parallel (Independ Sources) 33:00

Sample Problem: Series/Parallel (Independ Sources) 44:35

Sample Problem: Op Amps 49:10

Sample Problem: Op Amps 59:29

Sample Problem: Op Amps 65:33

Sample Problem: Op Amps 74:23

Sample Problem: Op Amps 86:22

Sample Problem: Op Amps 95:33

Sample Problem: Op Amps 109:17

Sample Problem: Op Amps 115:29

Sample Problem: Op Amps 124:59

教學方

Dr. Bonnie H. Ferri
Professor
Dr. Joyelle Harris
Director of the Engineering for Social Innovation Center
Dr Mary Ann Weitnauer

腳本

The topic of this problem is Kirchhoff's Current Law.

In this problem, we want to determine I1 and I2.

So we have a circuit that's drawn, and it has a voltage source in it.

It has four resistors and some currents that are already known.

The unknown currents are I1 and I2.

So to find these currents, we're going to use Kirchhoff's Current Law,

and we're going to sum the currents into the nodes.

So the first thing we have to do when we solve problems using

Kirchhoff's Current Law is we have to identify the nodes in the problem.

So we have a common node at the bottom.

We have a node on the left hand side of the circuit, and

we have a node on the right hand side of the circuit.

So there's three nodes associated with this circuit.

So if we want to use Kirchhoff's Current Law, we can sum the current into any one

of those nodes and see if we can solve for quantities that we have of interest.

So let's start with the node on the left hand side of the circuit.

So we want to sum the currents into the left hand node at the top.

And so if we do that, we see first of all, that we have I1 flowing into that node.

So it's I1 and

then we have three amps flowing out through the left most resistor.

So it's a -3 mA, and we also have I2,

which is flowing out of that part of the circuit as well.

And so that node has the equation as shown above,

so that is node 1 and we'll assume that is equation 1.

If we go to the other side of the circuit and

look at node 2, and we do the same thing.

We sum the currents into that node, then we see that I1 is flowing out, so

it's -I1, and we have 12mA that's flowing up into node 2 through the voltage source.

So it's +12 mA.

And we have 4 mA flowing down through and out of node 2 through

the resistor on the rightmost side of the circuit, so it's -4mA.

And that's equal to 0.

So, we can see that we can solve for I1, and then alpha mean for I2.

So, I1 in this case is going to be 8 mA,

and if I1 is 8 mA, then I2 using

equation 1 turns out to be 5 mA.

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